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Uncertainty in Future Events
The ability to forecast with high accuracy on future events is an important aspect of engineering. However, even with the best estimates, it is impossible to guarantee a 100% accuracy. Disruptive changes in technology, economy, and culture are usually bound to occur and alter the manner that future events occur. Specifically, disruptive changes can lead to a loss in value due to obsolescence, reduction in demand, or increase in operation cost (Sullivan, Wicks, & Koelling, 2014). In light of this, engineering economic analysis assesses the long-term consequences of projects while considering the time value of money.
Due to the inability to accurately foretell the future, engineers assess the cost and benefits of assets by comparing various scenarios such as most optimistic, most pessimistic, and most likely events that affect a product. They then give a weight to each scenario, which they later use to estimate the future cost or benefit of an event. A break-even analysis is conducted to examine the impact of the variability of estimates of the outcome (Newnan, Lavelle, & Eschenbach, 2013). It enables an engineer to estimate the amount of variability that a parameter may have before an initial decision is affected. However, the breakeven analysis does not show how an individual may take the inherent variability of parameters into account in an economic analysis (Blank & Tarquin, 2013). It is important to note that the salvage value is the component that mostly affects the differences between the economic alternatives. When different projects or assets have an almost similar salvage value, their economic differences may be small. On the contrary, a large difference in salvage values may make different projects to have huge economic differences.
The three range of scenarios; optimistic, pessimistic, and most likely are useful in determining the cost or benefits of an event. In a real-world scenario, an engineer is normally faced with a range of variables that fall within these categories. Ordinarily, most of these variables lie within the most likely range, while few of them are in the optimistic and pessimistic range (Newnan, Eschenbach, & Lavelle, 2004). The following equation, which considers the difference in weight of the variables is used to estimate the mean value of an asset.
Mean value= (optimistic+ (most likely)4+ pessimistic value))/6
Probability
Probabilities can also be used to determine the cost or benefits of an event. In this case, the weight of each class of costs or benefits is calculated using the probabilities. In probability, the sum of the probable ratios lies between zero and one. Although most research and analytical works use a lot of samples to estimate the most accurate probabilities, engineering normally uses between 2 and 5. The reason for the use of a small sample is because it is estimated by experts in their respective fields. In addition, each data requires detailed analysis to find if it is accurate (Newnan, Eschenbach, & Lavelle, 2004). Therefore, a small data compensates with the detailed analysis done on each sample.
When using probability to estimate the benefits or cost, the weight of each category, most likely, optimistic, and pessimistic are multiplied with the weights and then added up to estimate the overall cost or benefit of an event.
Joint Probability
In some cases, an asset or a project may have more than one probability due to various conditions that it must fulfill. For instance, a project may have the 0.2, 0.5, and 0.3 as the ratios for optimistic, most likely, and pessimistic respectively. Further, the asset may have a ratio of remaining in operation for 6 years as 0.7 and for 12 years as 0.3. This example represents a case of joint probability since each score (optimistic, most likely, and pessimistic) is joined with various probable scores of the time that the project may be in operation (6 years or 7 years). Moreover, both of these classes are independent of each other (Sullivan, Wicks, & Koelling, 2014).
The probability of A and B independent variable is presented as, P(A and B)= P(A) * P(B)
Expected Value
The expected value refers to the arithmetic mean that is calculated in probability distributions. The expected value is calculated by getting the sum of the weight of each scenario. The weights are calculated by multiplying each outcome with its probability of occurring (Blank & Tarquin, 2013).
Expected Value (A to N) = Outcome A* P(A)+ Outcome B*P(B)+… + Outcome N * P(N)
Economic Decision Trees
An economic decision tree refers to a tree like structure that is used to determine the probable outcomes of various decisions that engineers make. It usually has the following shape
 
D1
Decision node                                        D2      Decision maker chooses either of the available paths
 
Dx
 
 
 
 
 
p1       C1

 
 
Chance node                                         p2              C2   Represents probabilistic chance for each event.                                                                   p3
C3
Outcome (C1, C2,…,CY) Has probability (p1, p2, p3, …py) respectively.

 
 
Outcome node                                                shows the results of a particular path in a decision tree.
 
Pruned branch                          indicates that the branch has been pruned and another path has been chosen.
The chance nodes, decision node, and outcome nodes illustrate the problems structure.
Engineers make decisions of the best asset using the final nodes in a tree. The criteria used in the decision-making entails maximizing the present worth (PW) or minimizing the equivalent uniform annual cost (EUAC). The expected value for PW and EUAC is calculated at the chance nodes. A branch is chosen if it originates from the decision node that has the best PW or EUAC (Newnan, Lavelle, & Eschenbach, 2013). If it originates from a chance node, an individual calculates the expected value and inputs the value in the node. This process is important since it rolls back the values from the terminal nodes to the initial decision.
Risk
Risk refers to the probability that the company will get results that are not similar to the expected values, especially ones that lead to a loss or make the decision worse off. Risk is mainly measured using the probability of a loss. Alternatively, the standard deviation is used to measure how much the outcomes are dispersed from the expected values (Newnan, Eschenbach, & Lavelle, 2004). The standard deviation (α) is derived from the square root of the variance. The variance, on the other hand, refers to the difference between the mean and the actual values.
Standard deviation (α) = √ [EV(X – mean)2]
Standard deviation (α) = √ {EV(X2) – [EV(X)]2}
Standard deviation (α) = √ {Outcome2A x P (A) + Outcome2B x P(B)+… – expected value2}
Risk versus Return
The risk versus return refers to the probability of gaining in comparison to the probability of losing. A measure of the risk, which is calculated using standard deviation is placed on the x-axis where it is compared with the return, which is measured using the expected value that is placed on the y-axis. When assessing the projects to be considered, an individual checks for those that are not dominated by other projects (Newnan, Eschenbach, & Lavelle, 2004). Dominated projects are the ones that lie below the efficient frontier and have other projects above them in the chart. Projects that are in the efficient frontiers are the ones that are selected.
Simulation
Simulation refers to the process of assessing how an asset will be in its future time. Economic simulation uses random sampling of one of the variables to analyze an economic model for many iterations. The sampled values are used to calculate the PW, IRR, or EUAC for each iteration (Blank & Tarquin, 2013). The results from the iteration are then combined to create a probability distribution for the PW, IRR, or EUAC. Various analytical software such as Excel, @Risk, and Crystal Ball generate random numbers, which are used in simulation to assess possible future events.
Questions
10.1
Telephone poles exemplify items that have varying useful lives. Telephone poles, once installed in a location, remain in useful service until one of a variety of events occurs.

  • Name three reasons why a telephone pole might be removed from useful service at a particular location.

Solution
Damage by mites or lightning
Changes in technology resulting in the use of mobile phones instead of landline connections
Migration of individuals to various areas making the infrastructure unimportant
 
(b) You are to estimate the total useful life of telephone poles. If the pole is removed from an original location while it is still serviceable, it will be installed elsewhere. Estimate the optimistic life, most likely life, and pessimistic life for telephone poles. What percentage of all telephone poles would you expect to have a total useful life greater than your estimated optimistic life?
Solution
Optimistic Life is 25 years
Most likely life is 20 years
Pessimistic life is 10 years.
Total useful life= (optimistic+ (most likely)4+ pessimistic value))/6
Total useful life= (25+ (20* 4)+ 10)/6= 19.17 Years
10-6
Annual savings due to an energy efficiency project have a most likely value of $30,000. The high estimate of $40,000 has a probability of 0.2, and the low estimate of $20,000 has a probability of 0.30. What is the expected value for the annual savings?
Solution
Expected Value (A to N) = Outcome A* P(A)+ Outcome B*P(B)+… + Outcome N * P(N)
Sum of all probabilities = 1
Therefore, Most likely probability = 1- (0.2+ 0.3)= 0.5
Expected value = ((40,000* 0.2) + (30,000* 0.5) + (20,000* 0.3)
Expected value= 8,000+15,000+ 6,000= 29,000
10-15
A decision has been made to perform certain repairs on the outlet works of a small dam. For a particular 36-inch gate valve, there are three available alternatives:
(a) Leave the valve as it is.
(b) Repair the valve.
(c) Replace the valve.

  Leave Repair Replace Present Worth
Valve Seat 0.6 0.4 0.3 10,000
Valve Stem 0.5 0.3 0.2 20,000
Valve Body 0.4 0.2 0.1 30,000
Cost 0 10,000 20,000  

 
Loss due to leaving
0.6*10,000+ 0.5* 20,000+ 0.4* 30,000= 28,000
Cost of leaving is zero.
Net gain/ loss 28000+ 0= 28,000
Loss due to Repair
0.4*10,000+ 0.3* 20,000+ 0.2* 30,000= 16,000
Cost of Repair is 10,000
Net gain/ loss 16,000-+10,000= 26,000
Loss due to Replacement
0.3* 10,000+ 0.2* 20,000+ 0.1* 30,000= 10,000
Cost of Replacement= 20,000
Net gain/ loss 10,000+ 20,000= 30,000
Repairing is the best option. It results in the least expense for the dam owners, which is $26,000. Leaving the dam unrepaired costs $28,000 and replacing the valves costs $30,000.
10-24
Find the expected PW and probability distribution
First Cost        80000
Salvage value  0
Interest rate     9%
Shift/ day        Saving per year           Probability       Useful life       Probability
1                             15000                          0.3                   3                   0.6
2                             30000                           0.5                  5                    0.4
3                             45000                           0.2
Solution
PW (EV)= -80000 + (15,000 or 30000 or 45000){P/A 10%, 3 or 5}
P/A 9%, 3 year            P/A 9%, 5 year
2.5313                             3.88962
 

2.5313 3.88962
Annual benefit Probability Life Joint Probability PW PW* Joint Probability
15000 0.3 3 0.18 -42030.5 -7565.49
15000 0.3 5 0.12 -24966.7 -2996
30000 0.5 3 0.3 30088.6 9026.58
30000 0.5 5 0.2 33908.5 6781.7

 
10.25
Extend project life by 3 years at a cost of $50,000 at end of initial useful life.
Find the worth to the firm.
Solution
Present value of extension capital of 50000
PV= Capital* (1/(1+i)n
PV       After 3 years   After 5 years
38609.17         32496.57
PW (EV)= -118609.17 + (15,000 or 30000 or 45000){P/A, 9%, 6 years}
 
PW (EV)= -112496.57 + (15,000 or 30000 or 45000){P/A, 9%, 10 years}
 
P/A 9%, 6 year            P/A 9%, 8 year
4.4859                             5.5348                                                                                

Annual benefit Probability Life Joint Probability PW PW* Joint Probability
15000 0.3 6 0.18 -51320.7 -9237.72
15000 0.3 8 0.12 -29474.6 -3536.95
30000 0.5 6 0.3 15967.83 4790.349
30000 0.5 8 0.2 53547.43 10709.49
45000 0.2 6 0.12 83256.33 9990.76

 
 
 
 
 
 
 
 
 
 
References
Blank, L. & Tarquin, A. (2013). Basic engineering economy (2 Ed.). New York, NY: McGraw-Hill Education.
Newnan, D., Eschenbach, T., & Lavelle, J. (2004). Engineering economic analysis (9th Ed.). New York, NY: Oxford University Press.
Newnan, D., Lavelle, J., & Eschenbach, T. (2013). Engineering economic analysis (12th Ed.). New York, NY: Oxford University Press.
Sullivan, W., Wicks, E., & Koelling, P. (2014). Engineering economy (16th Ed.). Upper Saddle River, NJ: Person Publishers.